Special Relativity – Lin Jiahuang


Personally, I found (and still find) special relativity very, very counter-intuitive and had a hard time learning it. I suppose this is true for quite a handful of people, although there are some who told me that they find SR easy.

The main difficulty that I face is that of the inability to visualize a world with SR. I find it difficult to develop an intuition for SR, as I always try to do when I learn new topics. Personally, I find the development of intuition a very crucial step in gaining a deeper understanding of the topic. In this document, I will try to present some of the interesting ideas that may help us develop such an intuition. I will first show how the Lorentz transformation can be worked out. That will be a set of equations that will be very commonly used by many people doing SR. I will then apply those equations in a very interesting example. To those who have not studied SR, this example is full of “paradoxes”.

Lorentz Transformation

Lorentz Transformation is a set of equations that relates the space and time coordinates (x, y, z and t) of one frame to the space and time coordinates (x`, y`, z` and t`) of another frame of reference. If you are a robot, you will not see the world like how ordinary humans do. When there is a car crash 10m away from you at 10pm, you probably will not be able to tell that it is 10m away and the time is exactly 10pm. You only know that the event of the car crashing happens very near you late at night. However, the robot, with its ultrasonic sensor, is able to tell that the car crashed at a position exactly 10m away from it at 10pm. If the robot takes itself to be the origin, the robot will record the event to be happening at (10, 0, 0, 10). However, suppose that there is a robot between the car crash and the first robot, the recorded position will be (5, 0, 0, 10). This is the space and time coordinates of the other frame of reference.

Before Einstein, people used Galileans Relativity. The speed of light, c, has no special significance in it and things are allowed to travel faster than the speed of light. The length of a ruler measured by a person at rest with respect to the ruler and a person moving wrt the ruler is the same. Clocks run at the same rate to everyone. However, Maxwell’s Equations predicts that the speed of light is the same in all inertial reference frames. (This is one of the most commonly stated statements without proof when one studies SR. I have a way to work it out but the margin on this paper is too small for me to write it down.) Just this statement, plus the postulate that all laws of physics are valid in all inertial frames (not the best way to state it in an exam) results in all the rest of relativity.

Because of the two postulates, rulers appear shorter to people who are moving with respect to it according to the relation:, where gamma is known as the Lorentz factor.

Similarly, a clock that is stationary with respect to you will run faster compared to a clock that is moving. The relation is as such:.

These two effects are known as length contraction and time dilation respectively. It is important not to just remember the equations, but remember and visualize what exactly will happen in what kind of situation. This helps a lot when the situation gets more complicated and is an important step towards developing an intuition. Basically, just remember a moving ruler gets shorter (Do note that this applies for directions parallel to the moving frame only. The perpendicular lengths are not changed. To see why, imagine two dragons each on a train approaching each other. Each of them has the ability to blow out fire balls. Suppose that if the perpendicular length contracts, the fire ball of the “moving” dragon will not hit the other dragon. Yet, the situation is totally symmetrical. Hence, there will be a paradox.) and a moving clock runs slower.

Now, we can work out the Lorentz Transformation. We will work out the space transformation first. Imagine person A is standing at the train station while person is on an approaching train. Imagine when their origins coincide, they both start timing their clocks. At this instance, a bomb explodes 10m away in the positive x direction of the person on the platform. The person in the train will try to measure where the bomb explodes. One way to do so is to place rulers on the floor. To the person on the platform, he only needs to place 10 metre-rules from him to reach the bomb explosion point. However, the person the platform will see the person in the train placing more than 10 rulers on the floor, because the rulers are shortened. Each ruler will be shortened by the factor gamma. Hence, this gives the Lorentz transform for space:, where vt takes into account the distance the train moves if the event does not happen at the time the clock is started. Always remember that gamma is more than 1. In our example, t=0. Hence, x’ is larger than x. This means that the person in the carriage will measure a distance longer than the person on the platform, which is exactly the case if the person in the train places more rulers on the floor. Some people may try to use this equation to prove length contraction, only to realise that the primed x is actually larger than the un-primed x. It hence appears that length has not contracted, it has elongated. This way of proving is invalid because represents coordinates, not length of objects.

The next equation that we need to work out is the transformation for time. Before that, we need to introduce the idea of simultaneity. In the physical world, two events that are simultaneous to a person in one frame may not be simultaneous to another person in another inertial reference frame. Why is this so? Suppose that we want to synchronize two clocks, one of the easiest ways is to place a light source between the two clocks. The light source will send two pulses of light beams, one to each clock when it is activated. Since the distances between the clocks and the light source are the same, the two clocks will receive the signal at the same time. The clocks will start running when they receive the signal. Hence, the clocks are synchronized.

However, this is only true for the person at rest with respect to the clocks. Suppose that these clocks are in a moving train. Person A in the train will want to synchronise the clocks. Hence, he will perform what is stated above. To him, the clocks will indeed be synchronised. However, to a person outside the train, the clocks are not synchronised because suppose the train moves to the right, the light signal will take a longer time to reach the clock on the right because the light signal will need to travel a longer distance to reach the clock, since the train is also moving. On the other hand, the light signal will reach the back clock earlier because the clock is moving towards the light signal.

Suppose that the length of the train is L. Suppose that light takes time  to reach the clock 1 and  to reach clock 2. Suppose that the train travels at velocity v. Then,

for clock 1 and

for clock 2.

Hence, the difference in time is , which is


What does this time difference mean? It is the time difference between the activation of one clock and the other to the person on the platform. It is the time that the light signal takes to reach clock 1 and clock 2 has been activated.

Now, after this time difference, clock 1 will also be activated. However, what is the time on clock 2? After seconds after clock 2 is activated as measured by the person on the platform, clock 1 will be activated. When clock 1 is activated, due to time dilation, the reading on clock 2 is not , Instead, it is multiplied by , which is simply .

is a very important term that you should take note of. It will appear later and it reveals a very important piece of information: the difference in the time the clocks show in a moving frame with respect to you.

This is the intuition development part. For the case of length contraction, we can imagine that in the relativistic world, things that are moving with respect to you are squashed. In particular, rulers are squashed and hence measurements made will be different from your stationary frame. (If must be taken note that although the word “squashed” is used, you should not imagine the process of squashing. Instead, you should only see the result of squashing because suppose you want to squash a one dimensional world, there will always be a convergence point where things from its left will move towards the right and vice versa. However, in the real world, there is no such point. Many people when imagining the squashing process uses the origin of the coordinate system as the convergence point. However, this is not correct because once the origin of the coordinate system is changed, things will start to go wrong in your imagined world. No matter where the origin of the coordinate system is, the physical results will not changed because you are only adding a linear term into the Lorentz space transformation equation when your coordinate system is changed. )

So that is how you can imagine the relativistic world’s space. How about time? To imagine time, think of many imaginary clocks in the moving frame and your own stationary frame. The clocks in your frame are all synchronised. However, the clocks in the moving frame are all not synchronised. In fact, if the train moves towards the right, as with the previous example, two clocks a distance L apart (distance measured in their rest frame), will show a time difference of . In particular, if the train moves towards the right, the clock on the left will show a later time compared to clocks on the right. What do the readings on the clock mean? Suppose a bomb explodes at a certain position in space. This position has a certain x coordinate in your frame and another value of x in the moving frame’s coordinate. In your stationary frame, there are clocks all over, including at that position. Hence, to describe that event, you use the x position and the value on the clock at that position. This will be the description of the event from your reference frame. However, from the moving frame, the description will be based on another x coordinate and another value of time. This value of time is shown on the imaginary clock in the moving frame. Do note again that those imaginary clocks in the moving frame are all not synchronised.

With all these, we are now ready to look at how the time transformation equation can be worked out. Suppose that when the origin of the moving frame and your frame coincides, both of you synchronized and activates your clock. To you, all your clocks are synchronised. Suppose that the moving frame is moving towards the right side. All the clocks of the moving frame to the left  of the origin (the negative x direction) will be reading positive time and all the clocks to the right of the origin are all reading negative time (they are not activated yet). After a certain amount of time, an event happens at the t’at x’. What will be the description of the event by our resting frame? First of all, we know that when the description of the event is t’and x’, it means that the imaginary clock that is moving and not synchronised with all the surrounding clocks read t’. The position of the clock is x’. From this and the fact that two clocks a distance L have a time difference of , we will be able to find out the value of the clock at the origin. In this case, we know that L=x’. Hence, the value of the clock at the origin will be . However, we also know that due to time dilation, although the clock at the origin is synchronised with our resting clock, it does not have the same value as our resting clock after some time. In fact, the resting clock‘s value is multiplied by gamma and reads . With is, we have worked out the time transformation.

Interesting example

Suppose that there are two people, Jack and Jill. Jill is in a spacecraft travelling at the speed of 0.6c. There are two clocks that are spaced 6 light seconds apart.

At the instant when Jack meets Jill, both of their clocks are synchronised and started. Hence, Jill’s clock in the spaceship, Jack’s clock 1 and clock 2 are all started. Since the distance between the two clocks is 6 light seconds and Jill is travelling at 0.6c, Jack will measure Jill to take 10 seconds to reach clock 2, as measured by his clocks. Hence, both clocks 1 and 2 will show 10 s to Jack. However, Jill’s clock will read 8 s when she reaches clock 2 (go calculate yourself! Jill is travelling at 0.6c). So far so good, to Jack. Jack is happy because he sees the effect of time dilation and to him, there is no contradiction. However, what about Jill? Shouldn’t Jill also see time dilation? When Jill reaches clock 2, what does she see? She will see that her clock reads 8 s and clock 2 reads 10 s. You may ask, doesn’t this mean that to Jill, there is no time dilation? No! there is still time dilation and there is no contradiction. Jill, like Jack, is also able to observe time dilation. Indeed, clock 2 appears to read 10 s. However, that does not mean that 10 s has passed! The reason why 10 s has not passed is because clock 1 and 2 are in fact not synchronised. In fact, when clock 1 reads 0 (when they both touches their origin), clock 2 reads 3.6s . (Go work out yourself using  !) Hence, to Jill, 8 s has passed on her clock. However, only 6.4 seconds has passed on Jack’s watch. Hence, time dilation is still observed.

Now, I will leave the last part for you to think through yourself. Suppose that at the instance when both touches clock 2, both uses some super-pro telescope to look at clock 1. Jack will see the clock reading 4 s, since he knows that the clock that he is seeing is synchronised with clock 2 and clock 2 reads 10 s. Since clock 1 is 6 light seconds away, it has to read 4s. However, what does Jill see? She must also see 4 s. How can she reconcile the fact that she sees the clock to read 4 s when she knows that at the time she looks, the clock is reading 6.4 seconds? She can if she knows her relativity, using time dilation and length contraction. Since you know your relativity, try it!

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